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Potential problem

  1. Jun 18, 2013 #1
    1. The problem statement, all variables and given/known datais
    it is given in my highschool textbook that a positive charge is that higher potential than negative charge? i am not able to understant it.....can anyone please explain...

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 18, 2013 #2
    Look at your formula for potential. What happens if you replace q with -q?
     
    Last edited: Jun 18, 2013
  4. Jun 18, 2013 #3
    Start with the definition of electrostatic potential. What is it?
     
  5. Jun 18, 2013 #4
    well the definition says,"electrostatic potential at any point in a region of electrostatic field is the minimum work done in carrying a unit positive charge(without acceleration) from infinity to that point"
     
  6. Jun 18, 2013 #5
    Yes. Calculate the work done in bringing a unit positive charge from infinity to that point. Do you know the electric field produced by a point charge as a function of radius?
     
  7. Jun 18, 2013 #6
    calculating the work done in bringing a unit positive charge from infinity to that point in a region of electrostatic field will mean deriving formula formula for electrostatic potential due to a point charge......... so you mean to say what dreamlord said that is replace q with -q in the electrostatic potential formula due to a point charge....... electric field varies inversly to square of radius.....
     
  8. Jun 18, 2013 #7
    Yes, I am asking you to do what DimReg asked you to do in post #2, except derive it instead of using the formula.

    Another way of looking it is in terms of 'upstream' and 'downstream' field lines. Consider a unit positive charge in space. When you have a charge +q at the origin, you will have to apply a force to the test charge to move it 'upstream' against the field lines. This will raise the energy of the system. On the other hand, if you have a charge -q, you can simply leave the unit positive charge and it will move 'downstream' with the field lines. This will reduce the energy of the system as the test charge will gain kinetic energy.

    Basically, when there is repulsion, the potential energy increase, and when there is attraction, the energy decreases, because the system achieves a more stable state.
     
  9. Jun 18, 2013 #8
    so this will mean +16v will be greater than -18v even though magnitude wise 18 is greater than 16....
     
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